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Consider $f(x)$ function . We want to calculate $\lim_{x \to 3}f(x)$. So for left limit , we approach to $3$ and then compute $f(2.9) , f(2.99) , f(2.999)$ and so on . Now there is a weird thing . It is obvious that $2.9999.... = 3$ and also when we are talking about limit , point isn't important . In this case we don't take care about $f(3)$ but when we approach to $3$ infinitely , we get $3$ as $2.9999.... = 3$ ! . I'm very confused about these two concepts .

S.H.W
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2 Answers2

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No, in order to find that some real $l$ is the limit $$ \lim_{x\to3^-}f(x) $$ you don't compute $f(2.9)$, $f(2.99)$ and so on. And neither you compute $f(2.(9))$ (periodic $9$), because no assumption is made that $f$ is defined at $3$, nor the possible value of $f$ at $3$ is relevant for the existence of the limit.

Saying that $$ \lim_{x\to3^-}f(x)=l $$ means

for every $\varepsilon>0$ there exists $\delta>0$ such that, for $3-\delta<x<3$, it holds $|f(x)-l|<\varepsilon$.

You can compute $f(2.9...9)$, if you wish; it may give you an idea of what $l$ could be, but in general it won't.

egreg
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  • I'm sorry but I'm still confused . My problem is this : What's the difference between closing to a number infinitely and reaching it ? – S.H.W Apr 09 '17 at 16:07
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    @S.H.W There is no “closing to a number” in the definition I gave. Some teachers still like to use that language, notwithstanding that precise and rigorous definitions have been developed about 200 years ago. – egreg Apr 09 '17 at 16:16
  • Please take a look at it : https://www.mathsisfun.com/calculus/limits-formal.html – S.H.W Apr 09 '17 at 16:20
  • I think as you can see in "Mathisfun" page , the epsilon-delta definition is same as closing to a number . – S.H.W Apr 09 '17 at 16:32
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    @S.H.W I recommend you not to learn math from that site. – egreg Apr 09 '17 at 18:03
  • My calculus also said that. I think it is right. – S.H.W Apr 09 '17 at 18:42
  • Reviewing this question after 6 years, I think in addition to the provided answer this theorem helps to understand the convergence of a function at a point. So the convergence of $f(2.9),f(2,99),\dots$ isn't sufficient. For every sequence $a_n \to 3$ the sequence ${f(a_n)}_{n=1}^{\infty}$ should converge to $L$. – S.H.W Sep 25 '23 at 08:50
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    @S.H.W When I was a student of mathematics, our fellow engineering students all had powerful (for the time) calculators and the job of their instructors was inventing functions that would give hints to the actual limits only when going beyond the accuracy of those calculators. So, inputting $10^{-12}$ might have shown something near $3$, but the actual limit might have been $20$. And your remark is correct: you would need to check all sequences, not just one. This is why presenting “approaching” the way that site does is, to say the least, misleading. – egreg Sep 25 '23 at 09:16
  • Yes, now (although a little bit late): I understand your point. Thanks. – S.H.W Sep 25 '23 at 09:29
  • @S.H.W Don't worry! It took a couple of centuries to get calculus well founded! – egreg Sep 25 '23 at 09:31
  • Yes indeed :-). – S.H.W Sep 25 '23 at 09:38
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The idea of using a limit $\mathit{x}\rightarrow \mathit{n}$ is that you approach to $\mathit{n}$ as close as possible, but you actually never reach it.

Just forget that you are "computing" $f$ at every point because it is a missunderstanding. Imagine that you are moving along the graph of the function $f$, then when taking a limit you are getting as close as possible to a specific point without ever touching it, as the function does not need to be $defined$ at that point, or the image might be different than the limit itself.

Imagine the following case:

$$f(x) = \left \lbrace {x^2, x \not= 0 \atop 1 , x = 0}\right. $$

If you take $lim_{x\rightarrow0}f(x) = 0$ for both right and left limits, but the actual image is $f(0)=1$.

When one has the equallity between right limit, left limit and image at a certain point in a function, we then say that the function is $continuous$, but any function that is not continuous still has limits.

I hope I clarified that to you.

edit: keep in mind that when talking about real numbers, between any two numbers there is an infinity of more numbers, it doesn't matter how close you try to imagine them to be, and that is the idea exploited by the limit.

Ayeron
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  • I'm sorry but I'm still confused . My problem is this : What's the difference between closing to a number infinitely and reaching it ? – S.H.W Apr 09 '17 at 16:06
  • @S.H.W Ayeron gives a good example of the difference. As long as $x$ is not $0$ - no matter how close to $0$ it is then $f(x)=x^2$. But when $x=0$ ("reaching it" in your terms) $f(x)=1$. BTW, in standard real analysis $\epsilon, \delta$ are real numbers. Being real numbers they are not infinite, infinitesimals, or infinitely close to other real numbers. In other words there is a distinction between "arbitrarily close" and "infinitely close" or similar statements. There is a branch of math called "non-standard real analysis" that deals with infinitesimals. Wikipedia has an article on it. – Χpẘ Apr 17 '17 at 01:36
  • @user2460798 Consider this definition : $\forall \epsilon > 0 ; \exists ; \delta > 0: |x- 3| < \delta \Rightarrow |f(x) - L | < \epsilon$ . Can we choose $2.999...$ for $x$ ? – S.H.W Apr 17 '17 at 13:52
  • @user2460798 You said , "there is a distinction between "arbitrarily close" and "infinitely close" or similar statements." . In real analysis for finding the limit we are dealing with infinitely close or arbitrarily close to a number ? – S.H.W Apr 17 '17 at 13:57
  • Arbitrarily close. – Χpẘ Apr 17 '17 at 15:50
  • @S.H.W No you can't choose $2.999...$ because $\delta$ must be $>0$. – Χpẘ Apr 17 '17 at 15:55
  • @user2460798 Therefore if we compute finite numbers like ($f(2.9) , f(2.99) , .....)$ , how we can guarantee that its pattern will not change ? – S.H.W Apr 17 '17 at 19:26
  • You can't. You could have a term like $sin(1/(x-3))$ which changes sign more frequently the closer $x$ is to $3$. As has been said elsewhere in this thread, computing values using specific values of $x$ isn't how the delta/epsilon method works. – Χpẘ Apr 18 '17 at 00:58